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Codex  0 Created 2026-09-24 Updated 2026-09-25

y = \overset{x}{˙}

. Then

\overset{x}{˙} = y

\overset{y}{˙} = a + b x - y - x^{2} - y^{2}

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 30D / d / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

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The saddle contributes the full Gaussian

e^{i x} \int_{- \infty}^{\infty} e^{- x s^{2} /2} d s = 2 π, x^{- 1/2} e^{i x} .

(1)

Thus $α = 2 π$ . Since the phase is exactly quadratic, there are no further algebraic saddle terms; the next contribution is the endpoint term $O (x^{- 1} e^{- 3 x /2})$ .

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 30D / c / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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Deform the real ray upward to the saddle line, joining it near the endpoint and at a large right cutoff. The integrand is entire. The far bridge vanishes by Gaussian decay; the endpoint bridge is

O (x^{- 1} e^{- 3 x /2})

because

ℜ ϕ (- 1) = - 3/2

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 30D / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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The saddle is

z_{s} = 1 + i

. Completing the square gives

ϕ (z) = i - (z - 1 - i)^{2} /2

, so its steepest-descent line is the horizontal line

z = 1 + i + s

s \in R

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 30D / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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With phase

ϕ (z) = (1 + i) z - z^{2} /2

, the original ray is a contour integral. At the endpoint

z = - 1

, the steepest direction solves

Im (ϕ (z) - ϕ (- 1)) = 0

with decreasing real part; locally it is the direction for which

ϕ^{'} (- 1) (z + 1)

is negative real.

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 29L / e / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

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Let

k = N - n + 1

. The consumption first-order condition gives

C_{n}^{*} = \frac{X _{n - 1}^{*}}{k + 1} - \frac{k}{2} lo g γ = \frac{X _{n - 1}^{*}}{N - n + 2} - \frac{N - n + 1}{2} lo g γ .

(1)

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 29L / d / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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Starting from

A_{N} = 1

, recursion gives

A_{n} = 1/ (N - n + 1)

. The coefficient recursion gives

B_{n} = (N - n + 1) γ^{(N - n) /2}

, including

B_{N} = 1

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 29L / c / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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Substitution of

F (n + 1, x) = B_{n + 1} e^{- A_{n + 1} x}

separates the infima over investment and consumption. The investment contributes

γ = in f_{h} E (e^{- h R})

after rescaling, and the first-order condition preserves exponential form, with

A_{n} = A_{n + 1} / (1 + A_{n + 1})

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 29L / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

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With

G = - U = e^{- x}

and

F = - V

, maximization becomes

F (n, x) = c, h in f {G (c) + E [F (n + 1, x - c + h R)]}, F (N, x) = e^{- x} .

(1)

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 29L / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

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Let

V (n, x)

be maximal expected utility from dates

n + 1, \dots, N

plus terminal wealth, given

X_{n} = x

. Then

V (N, x) = U (x)

and

V (n, x) = c, h sup {U (c) + E [V (n + 1, x - c + h R_{n + 1})]} .

(1)

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 28L / d / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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Generate

B

resamples, compute

n (\hat{θ}^{* j} - \hat{θ})

, and use their empirical

α /2

and

1 - α /2

quantiles. Large

B

approximates the otherwise unavailable conditional bootstrap distribution.

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 28L / c / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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Replace the two

G

quantiles by conditional bootstrap quantiles

\overset{q}{^}_{u}

in the interval from (a). Uniform convergence implies

\overset{q}{^}_{u} \to q_{u}

in probability at continuity points, so Slutsky and continuity of

G

give coverage tending to

1 - α

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 28L / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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A bootstrap sample consists of

n

independent draws with replacement from the empirical distribution assigning mass

1/ n

to each observed

X_{i}

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 28L / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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Let

q_{u} = G^{- 1} (u)

. Since

n (\hat{θ} - θ) \sim G

, an exact interval is

[\hat{θ} - q_{1 - α /2} / n, \hat{θ} - q_{α /2} / n]

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 27K / d / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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ξ_{1} \leq C

, then

t < T_{N_{t} + 1} \leq t + C

. Taking expectations and using (b) yields

λ t - 1 < E N_{t} \leq λ (t + C) - 1

. Divide by

t

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 27K / c / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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The strong law gives

T_{n} / n \to 1/ λ

. Since

T_{N_{t}} \leq t < T_{N_{t} + 1}

and

N_{t} \to \infty

, sandwiching yields

N_{t} / t \to λ

almost surely.

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 27K / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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Apply part (a) to the stopping time

N_{t} + 1

to obtain

E T_{N_{t} + 1} = λ^{- 1} E (N_{t} + 1)

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 27K / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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This is Wald's identity. Expand the stopped sum and use

{M \geq i}

, which depends only on earlier interarrivals and is independent of

ξ_{i}

E \sum_{i = 1}^{M} ξ_{i} = \sum_{i} E [1_{M \geq i}] E ξ_{i} = λ^{- 1} EM

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 26L / d / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

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On disjoint events

A_{n}

with

P (A_{n}) = c / n^{2}

, set

X_{n} = n 1_{A_{n}}

. Then

X_{n} \to 0

almost surely and

E X_{n}^{2} = c

, but any common dominator has integral at least

\sum_{n} n P (A_{n}) = c \sum 1/ n = \infty

. Thus no integrable dominating variable exists.

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 26L / c / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

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Yes. The

L^{2}

bound makes the family uniformly integrable in

L^{1}

E [∣ X_{n} ∣ 1_{∣ X_{n} ∣ > K}] \leq E [X_{n}^{2}] / K \leq 1/ K

. Together with convergence in probability, Vitali's theorem gives

E ∣ X_{n} ∣ \to 0

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 Pinned article: Introduction to the OurBigBook Project

Welcome to the OurBigBook Project! Our goal is to create the perfect publishing platform for STEM subjects, and get university-level students to write the best free STEM tutorials ever.

Everyone is welcome to create an account and play with the site: ourbigbook.com/go/register. We belive that students themselves can write amazing tutorials, but teachers are welcome too. You can write about anything you want, it doesn't have to be STEM or even educational. Silly test content is very welcome and you won't be penalized in any way. Just keep it legal!

Video 1.

Intro to OurBigBook

. Source.

We have two killer features:

topics: topics group articles by different users with the same title, e.g. here is the topic for the "Fundamental Theorem of Calculus" ourbigbook.com/go/topic/fundamental-theorem-of-calculus
Articles of different users are sorted by upvote within each article page. This feature is a bit like:
- a Wikipedia where each user can have their own version of each article
- a Q&A website like Stack Overflow, where multiple people can give their views on a given topic, and the best ones are sorted by upvote. Except you don't need to wait for someone to ask first, and any topic goes, no matter how narrow or broad
This feature makes it possible for readers to find better explanations of any topic created by other writers. And it allows writers to create an explanation in a place that readers might actually find it.
Figure 1.
Screenshot of the "Derivative" topic page
. View it live at: ourbigbook.com/go/topic/derivative
Video 2.
OurBigBook Web topics demo
. Source.
local editing: you can store all your personal knowledge base content locally in a plaintext markup format that can be edited locally and published either:
- to OurBigBook.com to get awesome multi-user features like topics and likes
- as HTML files to a static website, which you can host yourself for free on many external providers like GitHub Pages, and remain in full control
This way you can be sure that even if OurBigBook.com were to go down one day (which we have no plans to do as it is quite cheap to host!), your content will still be perfectly readable as a static site.
Figure 2.
You can publish local OurBigBook lightweight markup files to either OurBigBook.com or as a static website
.
Figure 3.
Visual Studio Code extension installation
.
Figure 4.
Visual Studio Code extension tree navigation
.
Figure 5.
Web editor
. You can also edit articles on the Web editor without installing anything locally.
Video 3.
Edit locally and publish demo
. Source. This shows editing OurBigBook Markup and publishing it using the Visual Studio Code extension.
Video 4.
OurBigBook Visual Studio Code extension editing and navigation demo
. Source.
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Infinitely deep tables of contents:
Figure 6.
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.
Live URL: ourbigbook.com/cirosantilli/chordate
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